Conic Sections Question 106
Question: The equation of the common tangent touching the circle $ {{(x-3)}^{2}}+y^{2}=9 $ and the parabola $ y^{2}=4x $ above the x-axis, is
[IIT Screening 2001]
Options:
A) $ \sqrt{3}y=3x+1 $
B) $ \sqrt{3}y=-(x+3) $
C) $ \sqrt{3}y=x+3 $
D) $ \sqrt{3}y=-(3x+1) $
Show Answer
Answer:
Correct Answer: C
Solution:
Any tangent to $ y^{2}=4x $ is $ y=mx+\frac{1}{m}. $ It touches the circle, if $ 3=| \frac{3m+\frac{1}{m}}{\sqrt{1+m^{2}}} | $
or $ 9(1+m^{2})={{( 3m+\frac{1}{m} )}^{2}} $
or $ \frac{1}{m^{2}}=3 $ ,
$ \therefore m=\pm \frac{1}{\sqrt{3}}. $
For the common tangent to be above the x-axis, $ m=\frac{1}{\sqrt{3}} $
\Common tangent is, $ y=\frac{1}{\sqrt{3}}x+\sqrt{3} $
therefore $ \sqrt{3}y=x+3. $
BETA
